Construct homotopy from $(\alpha \cdot \beta) \cdot \gamma$ to $\alpha \cdot (\beta \cdot \gamma)$ explicitly I understand the general idea of a homotopy, but I'm a little lost on how to create them myself. For example if I wanted to show 
$$ \text{Let} \: \alpha, \beta, \text{and} \: \gamma \: \text{be paths} \: I \to X, \: \text{from} \: x_{0} \: \text{to} \: y_{0}, y_{0} \: \text{to} \: z_{0}, \: \text{and}  \: z_{0} \: \text{to} \: u_{0}. \: \text{Then} \: \\
(\alpha \cdot \beta) \cdot \gamma \sim \alpha \cdot (\beta \cdot \gamma) $$
A possible homotopy is $F: I \times I \to X$, given by 
$$\\ F(t,s) = 
\begin{cases}
\alpha(\frac{4t}{1+s}) & 0 \leq t \leq \frac{s+1}{4} \\
\beta(4t-1-s) & \frac{s+1}{4} \leq t \leq \frac{s+2}{4} \\
\gamma(\frac{4t - 2 - s}{2-s}) & \frac{s+2}{4} \leq t \leq 1 \\
\end{cases}$$
What I don't understand is where this comes from. What is the intuition here and how can I form explicit homotopies like this?
 A: See the picture below.

The idea is that, for any choice of $s$, the loop $t \mapsto F(t, s)$ consists of walking along $\alpha$, $\beta$ and $\gamma$, in that order. The difference between the various choices of $s$ is in the "time schedule": each choice of $s$ allocates different amounts of time to $\alpha$, $\beta$ and $\gamma$. 


*

*If $s = 0$, you walk path $\alpha$ in time $[0, \tfrac 1 4]$, then walk $\beta$ in time $[\tfrac 1 4, \tfrac 1 2 ]$, then walk $\gamma$ in time $[\tfrac 1 2 , 1]$.

*If $s = 1$, you walk path $\alpha$ in time $[0, \tfrac 1 2]$, then walk $\beta$ in time $[\tfrac 1 2, \tfrac 3 4]$, then walk $\gamma$ in time $[\tfrac 3 4 , 1]$.
For intermediate choices of $s$, the schedule is given by interpolation.
So for example, if $s = \tfrac 1 2$, you walk $\alpha$ in time $[0, \tfrac 3 8]$, then walk $\beta$ in time $[\tfrac 3 8, \tfrac 5 8]$, then walk $\gamma$ in time $[\tfrac 5 8, 1]$. And so on.
A: This is basically a reparametrization of the curve, but the parametrization is changing continuously. We may come up with the homotopy in two steps.


Let $u:I\to X$ be a curve, let $\phi:I\to I$ be a continuous map with $\phi(0)=0$ and $\phi(1)=1$, then $u$ is homotopic to $u\circ\phi$.


Proof: define $H(s, t)=u((1-s)t+s\phi(t))$, then $H(0,t)=u(t)$, $H(1,t)=u(\phi(t))$ and $H(s, 0)=u(0)$, $H(s, 1)=u(1)$. Essentially this is just mapping a homotopy between $\operatorname{Id}$ and $\phi$ in $I$ to the homotopy in $X$ by $u$.
We could construct $\phi:I\to I$ by rescaling the speed according to the time of travel through each curve. More explicitly, we have $[\alpha\cdot(\beta\cdot\gamma)] (t) = [(\alpha\cdot\beta) \cdot\gamma] (\phi(t)) $ if we define: $$\phi(t) =\begin{cases}\frac 12 t, & t\in[0,\frac 12]\\ t-\frac14, & t\in [\frac12, \frac34] \\ 2t-1, & t\in [\frac34, 1]\end{cases} $$
However if you apply the formula above, while we still get a homotopy, but the formula is not as clean (as in we need to divide into 5 cases). To obtain the formula you stated, let's look at the graphs of the intermediate reparametrizations, for each fixed $s_0$, the graph of $\phi_{s_0}(t) =1-s_0)t + s_0 \phi(t) $ would look like a "sheared up" version of $\phi(t)$. So image of the points at which $\phi_{s_0}$ are piecewisely defined do not match with the points at which $(\alpha\cdot\beta) \cdot\gamma$ is piecewisely defined.  To make them match we need to "shear" $\phi(t)$ to the left.
So we should take $\tilde\phi_s(t)$ the inverse function of $(1-s)t+s\phi^{-1}(t)$, i.e. $\tilde\phi_s(t)$ is the inverse of $$\begin{cases}(1+s)t, & t\in[0,\frac14] \\ t+\frac s4, & t\in[\frac 14,\frac 12]\\ \frac 12(2-s)t +\frac s2, & t\in[\frac 12,1]\end{cases}$$
Which is given by: $$\tilde\phi_s(t) =\begin{cases} \frac t{1+s}, & t\in[0,\frac {s+1}4]\\ t-\frac s4, & t\in[\frac{s+1}4,\frac{s+2}4]\\ \frac 2{2-s}t - \frac s{2-s}, & t\in[\frac {s+2}4,1]\end{cases} $$
The homotopy you provided in the post is then $H(s, t) =[(\alpha\cdot \beta)\cdot \gamma](\tilde\phi_s(t)) $. 
A: I have checked the definition and all axioms of the first homotopy group in this note and in this I also explain the exact homotopy you describe. I hope it helps you...
A: There is an easier way to do this by using the notion  of a "Moore path" as a pair $(f,r)$ where $r \in \mathbb  R, r \geqslant   0$ and $f:[0, \infty) \to X$ is constant $[r, \infty)$. This definition is used in the quite old book on Knot Theory by Crowell and Fox. Alternatively, as in Topology and Groupoids, one considers a path of length $r \geqslant 0$ to be a map $f:[0, r] \to X$. The composite of a path of length $r$ with a path of length $s$ is then, when defined, of length $r+s$. This corresponds intuitively to the idea of path as a "journey". Composition is then associative. We also have paths of length $0$ and in both definitions the composition of paths in $X$ gives a category $PX$. 
We also write $s$ for a  constant path of length $s$ at $y \in X$  and say two paths $f,g$ from $x$ to $y$ in $X$ are equivalent if there are real numbers $s,t \geqslant 0$ 
such that $f+s, g+t$ are homotopic rel end points. This gives the fundamental groupoid $\pi_1(X)$. 
One easily proves that any path is equivalent to a path of length $1$. This is called normalisation, which is not a process one uses for journeys. 
The formulae needed for all this work out much easier than the usual ones,  which seems to me a good thing.  (I've said all this somewhere else on this site.) 
